Determine the screw axis of a $3D$ rigid motion given by $F(x)=\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\\\end{pmatrix}x+\begin{pmatrix}1\\2\\3\\\end{pmatrix}$
Characteristic polynomial is $P(x)=-(\lambda-1)(\lambda+1)^2$
So I found the eigenvalues were $\lambda_1=1, \lambda_2=-1$ which had multiplicity $2$ for the matrix
Then the eigenvectors are $v_1=\begin{pmatrix}1\\1\\0\\\end{pmatrix}$
And $v_2=\begin{pmatrix}-1\\1\\0\\\end{pmatrix}$ and $v_3=\begin{pmatrix}0\\0\\1\\\end{pmatrix}$
I'm not really sure what to do from here, I thought the axis should correspond to one of these $3$ lines given by the eigenvectors, and that $\begin{pmatrix}1\\2\\3\\\end{pmatrix}$ should then lie on one of the lines. But that doesn't happen.
The determinant is $1$ so the matrix is a pure rotation, with no reflection.
The eigenvectors determine the reference system in which the action of the matrix is a stretching along the axes: in that reference the matrix becomes diagonal.
The product of the eigenvalues shall be $1$ as well. In this case they are $1, -1, -1$.
Clearly a vector along the rotation axis shall remain the same, and that is $(1,1,0)$ which is at $\pi /4$ in the $x,y$ plane.
The rotation is reversing the direction of the $z$ axis: so it is a rotation of $\pi$.
This will sweep the position of the $x$ and $y$ axes, which in fact is what the matrix does.
--- note in reply to your comment ---
A rigid rotation matrix must have at least one eigenvalue $=1$.
If it has two at $1$, then it must have all three ad it is a zero rotation matrix, otherwise the determinant is not one, and it is not a "pure" ("rigid body") rotation, but contains some stretching and/or reflection.